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Official PYQ · 2015Set 1 · Q53Transport Layer: Flow Control, Congestion Control, UDP, TCP, SocketsNAT1 mark

GATE CS 2015 Set 1 Q53 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Suppose that the stop-and-wait protocol is used on a link with a bit rate of 6464 kilobits\text{kilobits} per second and 2020 milliseconds\text{milliseconds} propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 5050 %\text{\%} is_________________.
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Answer: 320

For a Stop-and-Wait protocol, the link utilization (η\eta) is given by:

η=TtTt+2Tp\eta = \frac{T_t}{T_t + 2 T_p}

where:

  • TtT_t is the transmission delay of the frame: Tt=LBT_t = \frac{L}{B}
  • TpT_p is the propagation delay: Tp=20 ms=20×10−3 sT_p = 20\text{ ms} = 20 \times 10^{-3}\text{ s}
  • BB is the bandwidth: B=64 kbps=64×103 bpsB = 64\text{ kbps} = 64 \times 10^3\text{ bps}
  • LL is the frame size in bits

To achieve a link utilization of at least 50%50\% (η≥0.5\eta \ge 0.5):

TtTt+2Tp≥0.5  ⟹  Tt≥2Tp\frac{T_t}{T_t + 2 T_p} \ge 0.5 \implies T_t \ge 2 T_p

Substitute the values:

Tt≥2×20 ms=40 msT_t \ge 2 \times 20\text{ ms} = 40\text{ ms}

Now, calculate the minimum frame size LL in bits:

L64×103≥40×10−3  ⟹  L≥2560 bits\frac{L}{64 \times 10^3} \ge 40 \times 10^{-3} \implies L \ge 2560\text{ bits}

Converting the frame size to bytes:

Lbytes≥25608=320 bytesL_{\text{bytes}} \ge \frac{2560}{8} = 320\text{ bytes}

Thus, the minimum frame size is 320 bytes.

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